用Haskell寫一個傳遞閉包關系函式

我應該撰寫一個函式來回傳關系的傳遞閉包函式。這是我到目前為止所寫的內容：

relTrans :: [(Integer, Integer)] -> [(Integer, Integer)]
relTrans rel  |  rel == remove_dups(rel    relComp (remove_dups rel) (remove_dups rel))     =     rel
              |  otherwise                                                                  =     relTrans remove_dups(rel    relComp (remove_dups rel) (remove_dups rel))

這里提到的 remove_dups 和 relComp 函式如下 -

relComp :: [(Integer, Integer)] -> [(Integer, Integer)] -> [(Integer, Integer)]
relComp r1 r2 = [(a,c) | (a,k1)<-r1, (k2,c)<-r2, k1 == k2]

remove_dups :: [a] -> [a]
remove_dups [] = []
remove_dups (x:xs) = x : remove_dups (removeElem x XS)

removeElem :: a -> [a] -> [a]
removeElem n [] = []
removeElem n (m:zs) = if (n == m) then removeElem n zs else m:(removeElem n zs)

但是，我不斷收到以下錯誤：-

    * Couldn't match expected type `[(Integer, Integer)]
                                    -> [(Integer, Integer)]'
                  with actual type `[(Integer, Integer)]'
    * The function `relTrans' is applied to two arguments,
      but its type `[(Integer, Integer)] -> [(Integer, Integer)]'
      has only one
      In the expression:
        relTrans
          remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
      In an equation for `relTrans':
          relTrans rel
            | rel
                == remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
            = rel
            | otherwise
            = relTrans
                remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
    |       
183 |               |  otherwise                                                                  =     relTrans remove_dups(rel    relComp (remove_dups rel) (remove_dups rel))
    |                                                                                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

而這個錯誤——

    * Couldn't match expected type `[(Integer, Integer)]'
                  with actual type `[a0] -> [a0]'
    * Probable cause: `remove_dups' is applied to too few arguments
      In the first argument of `relTrans', namely `remove_dups'
      In the expression:
        relTrans
          remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
      In an equation for `relTrans':
          relTrans rel
            | rel
                == remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
            = rel
            | otherwise
            = relTrans
                remove_dups (rel    relComp (remove_dups rel) (remove_dups rel))
    |       
183 |               |  otherwise                                                                  =     relTrans remove_dups(rel    relComp (remove_dups rel) (remove_dups rel))
    |                                                                                                            ^^^^^^^^^^^

有人可以幫我解決為什么我收到這些錯誤嗎？

uj5u.com熱心網友回復：

您似乎在這里嘗試使用 C 風格的函式應用程式：

  relTrans remove_dups(rel    relComp (remove_dups rel) (remove_dups rel))
  --             here ^

在 Haskell 中，將relTrans函式應用于兩個引數：remove_dupsand (rel ...)，但relTrans只有一個引數，所以你會得到一個錯誤。

你可能打算這樣寫：

  relTrans (remove_dups (rel    relComp (remove_dups rel) (remove_dups rel)))

標籤：哈斯克尔 函数式编程

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